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If $u$ is harmonic and bounded in $\mathbb{C}$, then I've shown that $u$ is constant. I guess it can be helpful to show that $u \equiv 0$ if $u$ is harmonic and $\lim_{|z| \to \infty} u(z) = 0$, but how? I guess $u$ is not neccesary bounded in $\mathbb{C}$. What I can get is for all $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that $|u(z)| < \varepsilon$ if $|z| > N$. Any help is appreciated, thanks

joseabp91
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    Harmonic function attain maximum on boundary. Hence, if $|u(z)| < \epsilon$ on a disk boundary with radius $R$ is is bounded by the same value on the whole disk. – Salcio May 04 '21 at 22:42
  • Adding to the previous comment, note that harmonic functions are in particular continuous and continuous functions on compact sets are bounded (since the image of a compact set under a continuous function is compact). So if $|u(z)| < \varepsilon$ for any $|z| < N$, then noting also that $u$ is bounded on the closed ball $B(0, N)$ we conclude that $u$ is bounded. – Quoka May 04 '21 at 22:45
  • continuous functions on compact sets are bounded, but they are bounded in the compact set, right? I mean it is not bounded in $\mathbb{C}$ neccesarily – joseabp91 May 04 '21 at 22:51
  • Using $\epsilon=1$ in the definition of limits, there exists $N$ such that for all $|z|>N$, we have $|u(z)-0|<1$. Next, $u$ being continuous on the compact disc ${|\zeta|\leq N}$ implies it is bounded here, say by a bound $B$. So, for all $z\in\Bbb{C}$, we have $|u(z)|\leq B+1$ (because $u(z)$ is bounded by $B$ if $|z|\leq N$ and is bounded by $1$ if $|z|>N$). – peek-a-boo May 04 '21 at 22:58
  • I see thank you very much – joseabp91 May 04 '21 at 23:05

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